No group of order 10000 is simple
$begingroup$
A proof of this fact was already given here: No group of order 10,000 is simple
However, I am wondering whether or not the following proof works as well:
By way of contradiction, suppose $G$ is simple and $|G| = 10000 = 5^42^4$. Sylow theory gives $|operatorname{Syl}_5(G)| = 1$ or $16$. If $|operatorname{Syl}_5(G)| = 1$, then there is a Sylow 5-subgroup that is normal, and so we would have a contradiction. So suppose that $|operatorname{Syl}_5(G)| = 16$. Consider the action of $G$ on $operatorname{Syl}_5(G)$ by conjugation and let $$phi : G to S_{16}$$ be the associated permutation representation. The map $phi$ is nontrivial since the action is transitive by the second part of Sylow theory, which says that all Sylow p-subgroups are conjugate of each other. This show that the kernel of $phi$ is not all of $G$. Also, note that $10^4 = 5^4times 2^4$ does not divide $16!$, since $16! = 2^{15} times 3^6times 5^3times 7^2times 11times 13$, and this prime factorization does not contain $5^4$. Hence $phi$ is not injective, and so the kernel is not trivial. Hence $ker(phi)$ is a proper nontrivial normal subgroup of $G$, which contradicts out assumption that $G$ is simple.
group-theory proof-verification sylow-theory simple-groups
edited Jan 18 at 16:22
Rei Henigman
717
asked Dec 8 '18 at 23:20
WesleyWesley
518313
$endgroup$
add a comment |
$begingroup$
A proof of this fact was already given here: No group of order 10,000 is simple
However, I am wondering whether or not the following proof works as well:
By way of contradiction, suppose $G$ is simple and $|G| = 10000 = 5^42^4$. Sylow theory gives $|operatorname{Syl}_5(G)| = 1$ or $16$. If $|operatorname{Syl}_5(G)| = 1$, then there is a Sylow 5-subgroup that is normal, and so we would have a contradiction. So suppose that $|operatorname{Syl}_5(G)| = 16$. Consider the action of $G$ on $operatorname{Syl}_5(G)$ by conjugation and let $$phi : G to S_{16}$$ be the associated permutation representation. The map $phi$ is nontrivial since the action is transitive by the second part of Sylow theory, which says that all Sylow p-subgroups are conjugate of each other. This show that the kernel of $phi$ is not all of $G$. Also, note that $10^4 = 5^4times 2^4$ does not divide $16!$, since $16! = 2^{15} times 3^6times 5^3times 7^2times 11times 13$, and this prime factorization does not contain $5^4$. Hence $phi$ is not injective, and so the kernel is not trivial. Hence $ker(phi)$ is a proper nontrivial normal subgroup of $G$, which contradicts out assumption that $G$ is simple.
group-theory proof-verification sylow-theory simple-groups
edited Jan 18 at 16:22
Rei Henigman
717
asked Dec 8 '18 at 23:20
WesleyWesley
518313
$endgroup$
1
$begingroup$
I might be missing something, but this is the argument given in brief by the accepted solution to the linked problem. e: Here the action on left cosets of the normalizer of a given sylow 2-subgroup is quite similar to the conjugation action on the set of sylow 2-subgroups
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 23:23
1
$begingroup$
This is not a proof by contradiction: you have two possible cases, for which you find a non-trivial normal subgroup, hence in each case $G$ is not simple.
$endgroup$
– Bernard
Dec 8 '18 at 23:24
$begingroup$
@RolfHoyer So this a valid and complete proof, even though I am using a different action than the other proof?
$endgroup$
– Wesley
Dec 9 '18 at 20:11
add a comment |
$begingroup$
A proof of this fact was already given here: No group of order 10,000 is simple
However, I am wondering whether or not the following proof works as well:
By way of contradiction, suppose $G$ is simple and $|G| = 10000 = 5^42^4$. Sylow theory gives $|operatorname{Syl}_5(G)| = 1$ or $16$. If $|operatorname{Syl}_5(G)| = 1$, then there is a Sylow 5-subgroup that is normal, and so we would have a contradiction. So suppose that $|operatorname{Syl}_5(G)| = 16$. Consider the action of $G$ on $operatorname{Syl}_5(G)$ by conjugation and let $$phi : G to S_{16}$$ be the associated permutation representation. The map $phi$ is nontrivial since the action is transitive by the second part of Sylow theory, which says that all Sylow p-subgroups are conjugate of each other. This show that the kernel of $phi$ is not all of $G$. Also, note that $10^4 = 5^4times 2^4$ does not divide $16!$, since $16! = 2^{15} times 3^6times 5^3times 7^2times 11times 13$, and this prime factorization does not contain $5^4$. Hence $phi$ is not injective, and so the kernel is not trivial. Hence $ker(phi)$ is a proper nontrivial normal subgroup of $G$, which contradicts out assumption that $G$ is simple.
group-theory proof-verification sylow-theory simple-groups
edited Jan 18 at 16:22
Rei Henigman
717
asked Dec 8 '18 at 23:20
WesleyWesley
518313
$endgroup$
A proof of this fact was already given here: No group of order 10,000 is simple
However, I am wondering whether or not the following proof works as well:
By way of contradiction, suppose $G$ is simple and $|G| = 10000 = 5^42^4$. Sylow theory gives $|operatorname{Syl}_5(G)| = 1$ or $16$. If $|operatorname{Syl}_5(G)| = 1$, then there is a Sylow 5-subgroup that is normal, and so we would have a contradiction. So suppose that $|operatorname{Syl}_5(G)| = 16$. Consider the action of $G$ on $operatorname{Syl}_5(G)$ by conjugation and let $$phi : G to S_{16}$$ be the associated permutation representation. The map $phi$ is nontrivial since the action is transitive by the second part of Sylow theory, which says that all Sylow p-subgroups are conjugate of each other. This show that the kernel of $phi$ is not all of $G$. Also, note that $10^4 = 5^4times 2^4$ does not divide $16!$, since $16! = 2^{15} times 3^6times 5^3times 7^2times 11times 13$, and this prime factorization does not contain $5^4$. Hence $phi$ is not injective, and so the kernel is not trivial. Hence $ker(phi)$ is a proper nontrivial normal subgroup of $G$, which contradicts out assumption that $G$ is simple.
group-theory proof-verification sylow-theory simple-groups
group-theory proof-verification sylow-theory simple-groups
edited Jan 18 at 16:22
Rei Henigman
717
asked Dec 8 '18 at 23:20
WesleyWesley
518313
edited Jan 18 at 16:22
Rei Henigman
717
asked Dec 8 '18 at 23:20
WesleyWesley
518313
edited Jan 18 at 16:22
Rei Henigman
717
edited Jan 18 at 16:22
Rei Henigman
717
edited Jan 18 at 16:22
Rei Henigman
717
717
asked Dec 8 '18 at 23:20
WesleyWesley
518313
asked Dec 8 '18 at 23:20
WesleyWesley
518313
asked Dec 8 '18 at 23:20
WesleyWesley
518313
518313
1
$begingroup$
I might be missing something, but this is the argument given in brief by the accepted solution to the linked problem. e: Here the action on left cosets of the normalizer of a given sylow 2-subgroup is quite similar to the conjugation action on the set of sylow 2-subgroups
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 23:23
1
$begingroup$
This is not a proof by contradiction: you have two possible cases, for which you find a non-trivial normal subgroup, hence in each case $G$ is not simple.
$endgroup$
– Bernard
Dec 8 '18 at 23:24
$begingroup$
@RolfHoyer So this a valid and complete proof, even though I am using a different action than the other proof?
$endgroup$
– Wesley
Dec 9 '18 at 20:11
add a comment |
1
$begingroup$
I might be missing something, but this is the argument given in brief by the accepted solution to the linked problem. e: Here the action on left cosets of the normalizer of a given sylow 2-subgroup is quite similar to the conjugation action on the set of sylow 2-subgroups
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 23:23
1
$begingroup$
This is not a proof by contradiction: you have two possible cases, for which you find a non-trivial normal subgroup, hence in each case $G$ is not simple.
$endgroup$
– Bernard
Dec 8 '18 at 23:24
$begingroup$
@RolfHoyer So this a valid and complete proof, even though I am using a different action than the other proof?
$endgroup$
– Wesley
Dec 9 '18 at 20:11
1
1
$begingroup$
I might be missing something, but this is the argument given in brief by the accepted solution to the linked problem. e: Here the action on left cosets of the normalizer of a given sylow 2-subgroup is quite similar to the conjugation action on the set of sylow 2-subgroups
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 23:23
$begingroup$
I might be missing something, but this is the argument given in brief by the accepted solution to the linked problem. e: Here the action on left cosets of the normalizer of a given sylow 2-subgroup is quite similar to the conjugation action on the set of sylow 2-subgroups
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 23:23
1
1
$begingroup$
This is not a proof by contradiction: you have two possible cases, for which you find a non-trivial normal subgroup, hence in each case $G$ is not simple.
$endgroup$
– Bernard
Dec 8 '18 at 23:24
$begingroup$
This is not a proof by contradiction: you have two possible cases, for which you find a non-trivial normal subgroup, hence in each case $G$ is not simple.
$endgroup$
– Bernard
Dec 8 '18 at 23:24
$begingroup$
@RolfHoyer So this a valid and complete proof, even though I am using a different action than the other proof?
$endgroup$
– Wesley
Dec 9 '18 at 20:11
$begingroup$
@RolfHoyer So this a valid and complete proof, even though I am using a different action than the other proof?
$endgroup$
– Wesley
Dec 9 '18 at 20:11
add a comment |
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$begingroup$
I might be missing something, but this is the argument given in brief by the accepted solution to the linked problem. e: Here the action on left cosets of the normalizer of a given sylow 2-subgroup is quite similar to the conjugation action on the set of sylow 2-subgroups
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 23:23
1
$begingroup$
This is not a proof by contradiction: you have two possible cases, for which you find a non-trivial normal subgroup, hence in each case $G$ is not simple.
$endgroup$
– Bernard
Dec 8 '18 at 23:24
$begingroup$
@RolfHoyer So this a valid and complete proof, even though I am using a different action than the other proof?
$endgroup$
– Wesley
Dec 9 '18 at 20:11